Fluidos

Chapter 2 • Pressure Distribution in a Fluid
2.1 For the two-dimensional stress field in Fig. P2.1, let

σ xx = 3000 psf σ yy = 2000 psf σ xy = 500 psf
Find the shear and normal stresses on plane AA cutting through at 30°. Solution: Make cut “AA” so that it just hits the bottom right corner of the element. This gives the freebody shown at right. Now sum forces normal and tangential to sideAA. Denote side length AA as “L.”

Fig. P2.1

å Fn,AA = 0 = σ AA L − (3000 sin 30 + 500 cos30)L sin 30 − (2000 cos 30 + 500 sin 30)L cos 30
Solve for σ AA ≈ 2683 lbf/ft 2 Ans. (a)

å Ft,AA = 0 = τ AA L − (3000 cos30 − 500 sin 30)L sin 30 − (500 cos30 − 2000 sin 30)L cos30 Solve for τ AA ≈ 683 lbf/ft 2 Ans. (b)

2.2 For the stress field of Fig. P2.1, change the known data to σxx = 2000 psf,σyy = 3000 psf, and σn(AA) = 2500 psf. Compute σxy and the shear stress on plane AA. Solution: Sum forces normal to and tangential to AA in the element freebody above, with σn(AA) known and σxy unknown:
å Fn,AA = 2500L − (σ xy cos30° + 2000 sin 30°)L sin 30° − (σ xy sin 30° + 3000 cos 30°)L cos30° = 0

Solve for σ xy = (2500 − 500 − 2250)/0.866 ≈ − 289 lbf/ft 2

Ans. (a)

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SolutionsManual • Fluid Mechanics, Fifth Edition

In like manner, solve for the shear stress on plane AA, using our result for σxy:

å Ft,AA = τ AA L − (2000 cos30 ° + 289sin 30°)L sin 30° + (289 cos30° + 3000 sin 30°)L cos30° = 0
Solve for τ AA = 938 − 1515 ≈ − 577 lbf/ft 2
Ans. (b)

This problem and Prob. 2.1 can also be solved using Mohr’s circle.

2.3 A vertical clean glass piezometer tube hasan inside diameter of 1 mm. When a pressure is applied, water at 20°C rises into the tube to a height of 25 cm. After correcting for surface tension, estimate the applied pressure in Pa. Solution: For water, let Y = 0.073 N/m, contact angle θ = 0°, and γ = 9790 N/m3. The capillary rise in the tube, from Example 1.9 of the text, is
hcap =
2Y cosθ 2(0.073 N /m) cos(0°) = = 0.030 m γR (9790 N /m3)(0.0005 m)

Then the rise due to applied pressure is less by that amount: hpress = 0.25 m − 0.03 m = 0.22 m. The applied pressure is estimated to be p = γhpress = (9790 N/m3)(0.22 m) ≈ 2160 Pa Ans.
2.4 Given a flow pattern with isobars po − Bz + Cx2 = constant. Find an expression x = fcn(z) for the family of lines everywhere parallel to the local pressure gradient ∇p. Solution: Find the slope(dx/dz) of the isobars and take the negative inverse and integrate:
d dx dx −1 |p =const = B = (p o − Bz + Cx 2 ) = − B + 2Cx = 0, or: dz dz dz 2Cx (dx/dz)gradient Thus dx |gradient = − 2Cx , integrate ò dx = ò −2C dz , x = const e −2Cz/B dz B x B Ans.

2.5 Atlanta, Georgia, has an average altitude of 1100 ft. On a U.S. standard day, pressure gage A reads 93 kPa and gage B reads 105 kPa. Expressthese readings in gage or vacuum pressure, whichever is appropriate.

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Solution: We can find atmospheric pressure by either interpolating in Appendix Table A.6 or, more accurately, evaluate Eq. (2.27) at 1100 ft ≈ 335 m:

æ Bz ö pa = po ç 1 − ÷ To ø è Therefore:

g/RB

é (0.0065 K/m)(335 m) ù = (101.35 kPa) ê1 − ú 288.16 K ë û

5.26≈ 97.4 kPa

Gage A = 93 kPa − 97.4 kPa = −4.4 kPa (gage) = + 4.4 kPa (vacuum) Gage B = 105 kPa − 97.4 kPa = + 7.6 kPa (gage) Ans.
2.6 Express standard atmospheric pressure as a head, h = p/ρ g, in (a) feet of ethylene glycol; (b) inches of mercury; (c) meters of water; and (d) mm of methanol. Solution: Take the specific weights, γ = ρ g, from Table A.3, divide patm by γ :

(a) Ethyleneglycol: h = (2116 lbf/ft2)/(69.7 lbf/ft3) ≈ 30.3 ft Ans. (a) (b) Mercury: h = (2116 lbf/ft2)/(846 lbf/ft3) = 2.50 ft ≈ 30.0 inches Ans. (b) (c) Water: h = (101350 N/m2)/(9790 N/m3) ≈ 10.35 m
Ans. (c)

(d) Methanol: h = (101350 N/m2)/(7760 N/m3) = 13.1 m ≈ 13100 mm Ans. (d)

2.7 The deepest point in the ocean is 11034 m in the Mariana Tranch in the Pacific. At this depth γseawater ≈ 10520...